Hello! I don't know how to foil and add/subtract an imaginary numbers.

$\displaystyle x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i)) $

thanks...

Printable View

- Mar 14th 2010, 06:02 PMAnemoriimaginary numbers
Hello! I don't know how to foil and add/subtract an imaginary numbers.

$\displaystyle x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i)) $

thanks... - Mar 14th 2010, 06:27 PMskeeter
- Mar 14th 2010, 07:03 PMAnemori
- Mar 14th 2010, 09:29 PMAnemori
- Mar 15th 2010, 10:45 AMskeeter
- Mar 15th 2010, 12:56 PMAnemori
- Mar 15th 2010, 01:14 PMskeeter
- Mar 15th 2010, 03:56 PMAnemori
- Mar 16th 2010, 09:33 PMAnemori
- Mar 17th 2010, 06:29 AMskeeter
is that so?

$\displaystyle f(2i) = (2i)^6 - 6(2i)^5 + 14(2i)^4 - 24(2i)^3 + 40(2i)^2

$

$\displaystyle f(2i) = 64i^6 - 192i^5 + 224i^4 - 192i^3 + 160i^2$

$\displaystyle f(2i) = -64 - 192i + 224 + 192i - 160 = 0$

... and I guarantee you that $\displaystyle f(3+i) = 0$ ... complex zeros always come in conjugate pairs. - Mar 17th 2010, 07:15 AMChris L T521
There is enough information for the user to solve the problem. There is no need to carry on this discussion anymore.

Thread closed.